we have a gaseous mixture @ 15 degree Celsius at 1 atm and its entering at a rat

Question

we have a gaseous mixture @ 15 degree Celsius at 1 atm and its entering at a rate of change of 2.5 m^3/min. The mixture has the following composition:

30% molar Methane    (mm=16.04 g/mol)

10% molar Ethane      (mm= 30.07 g/mol)

Molecular Hidrogen      (mm=3.03 g/mol)

Please find:

-Molarity

-molar Mass of mixture

- Density of mixture

-specific gravity of mixture (you may use the specific gravity of air as a reference 1.29 kg/m^3)

- rate of masic flow (lbm/hr)

-masic fraction of components

- volumetric fraction

PLEASE SHOW PROCEDURE :)

Explanation / Answer

Volume of gas mixture entering per minute = 2.5 m3 = 2500 litres

Moles of gas mixture entering, n = (P*V)/(R*T) = (1*2500)/(0.0821*288) = 105.73

Now, moles of methane = 0.3*105.73 = 31.719

mass of ethane entering = moles*molar mass = 31.719*16.04 = 508.773 g

moles of ethane = 0.1*105.73 = 10.573

mass of ethane entering = moles*molar mass = 10.573*30.07 = 317.93 g

moles of hydrogen = 0.6*105.73 = 63.438

mass of hydrogen entering =moles*molar mass = 63.438*3.03 = 192.217 g

Thus, total mass of gas entering = 1018.92 g

Molar mass of the gas mixture = mass of the gas mixture/moles of the gas mixture = 9.637 g/mole

density of the gas mixture = mass/volume = 1018.92/2500 = 0.408 g/L

specific gravity = density of the gas mixture/density of air = 0.408/1.29 = 0.316

Molarity of thegas mixture =moles/volume of solution in litres = 105.73/2500 = 0.0423 M

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